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401. The parallel axis theorem is used to calculate the moment of inertia of a body:
ⓐ. Only about the centroidal axis
ⓑ. About any axis parallel to the centroidal axis
ⓒ. Only about an axis at the edge of the body
ⓓ. Only for spherical bodies
Correct Answer: About any axis parallel to the centroidal axis
Explanation: The parallel axis theorem allows calculation of the moment of inertia about any axis parallel to the centroidal axis using the known moment of inertia about the centroidal axis.
402. The mathematical statement of the parallel axis theorem is:
ⓐ. \(I = I_{\text{cm}} – Md^2\)
ⓑ. \(I = I_{\text{cm}} + Md^2\)
ⓒ. \(I = I_{\text{cm}} \times Md^2\)
ⓓ. \(I = \frac{I_{\text{cm}}}{Md^2}\)
Correct Answer: \(I = I_{\text{cm}} + Md^2\)
Explanation: According to the parallel axis theorem, \(I\) about any axis parallel to the centroidal axis is the sum of the moment of inertia about the centroidal axis \(I_{\text{cm}}\) and \(Md^2\), where \(d\) is the distance between the two axes.
403. In the parallel axis theorem formula \(I = I_{\text{cm}} + Md^2\), what does \(d\) represent?
ⓐ. Diameter of the body
ⓑ. Distance between the centroidal axis and the given axis
ⓒ. Radius of gyration
ⓓ. Distance between two particles
Correct Answer: Distance between the centroidal axis and the given axis
Explanation: Here, \(d\) is the perpendicular distance between the axis through the center of mass and the parallel axis about which we want the moment of inertia.
404. The parallel axis theorem is valid for:
ⓐ. Only one-dimensional bodies
ⓑ. Any rigid body
ⓒ. Only circular laminae
ⓓ. Only rods and discs
Correct Answer: Any rigid body
Explanation: The theorem applies to any rigid body provided the new axis is parallel to the centroidal axis and lies at a finite distance \(d\).
405. Which of the following is a direct application of the parallel axis theorem?
ⓐ. Moment of inertia of a disc about its central axis
ⓑ. Moment of inertia of a rod about an axis through its midpoint
ⓒ. Moment of inertia of a rod about an axis through one end, perpendicular to its length
ⓓ. Moment of inertia of a sphere about its diameter
Correct Answer: Moment of inertia of a rod about an axis through one end, perpendicular to its length
Explanation: Using the theorem: \(I = I_{\text{cm}} + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2\).
406. The parallel axis theorem was derived using which principle?
ⓐ. Law of gravitation
ⓑ. Definition of torque
ⓒ. Definition of moment of inertia with respect to point masses
ⓓ. Hooke’s law
Correct Answer: Definition of moment of inertia with respect to point masses
Explanation: The proof of the theorem expands \(r^2 = (x + d)^2\) for particles, showing that the new moment of inertia is \(I_{\text{cm}} + Md^2\).
407. Which of the following steps is essential in the proof of the parallel axis theorem?
ⓐ. Considering the body as a system of point masses
ⓑ. Assuming the axis passes through the surface only
ⓒ. Neglecting the mass of the body
ⓓ. Using torque equation directly
Correct Answer: Considering the body as a system of point masses
Explanation: The proof begins by expressing moment of inertia as \(I = \sum m_i r_i^2\), where distances are redefined relative to the new axis.
408. The term \(Md^2\) in the parallel axis theorem represents:
ⓐ. Additional torque
ⓑ. Increase in mass of the system
ⓒ. Additional moment of inertia due to shifting of axis
ⓓ. Change in radius of gyration
Correct Answer: Additional moment of inertia due to shifting of axis
Explanation: \(Md^2\) accounts for the increased rotational inertia because the axis is shifted away from the centroid by distance \(d\).
409. What is the moment of inertia of a solid sphere of mass \(M\) and radius \(R\) about a tangent to its surface?
ⓐ. \(\frac{2}{5}MR^2\)
ⓑ. \(\frac{7}{5}MR^2\)
ⓒ. \(\frac{5}{7}MR^2\)
ⓓ. \(\frac{9}{5}MR^2\)
Correct Answer: \(\frac{7}{5}MR^2\)
Explanation: By parallel axis theorem: \(I = I_{\text{cm}} + Md^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2\).
410. The parallel axis theorem is particularly useful when:
ⓐ. The axis passes through the center of mass
ⓑ. The axis passes through the edge or tangent of the body
ⓒ. The axis is inside the plane of lamina
ⓓ. The axis is not parallel to the centroidal axis
Correct Answer: The axis passes through the edge or tangent of the body
Explanation: In practical cases, such as calculating the moment of inertia of rods, discs, and spheres about axes at the edge, the parallel axis theorem provides a quick way to compute values without redoing the integration.
411. The perpendicular axis theorem applies to which type of bodies?
ⓐ. One-dimensional rigid bodies
ⓑ. Three-dimensional solid bodies
ⓒ. Planar laminae (two-dimensional bodies)
ⓓ. Fluids in rotation
Correct Answer: Planar laminae (two-dimensional bodies)
Explanation: The perpendicular axis theorem applies only to flat laminae lying in a plane. It relates the moment of inertia about an axis perpendicular to the plane to the moments of inertia about two perpendicular axes lying in the plane.
412. The mathematical statement of the perpendicular axis theorem is:
ⓐ. \(I_z = I_x – I_y\)
ⓑ. \(I_z = I_x + I_y\)
ⓒ. \(I_z = I_x \times I_y\)
ⓓ. \(I_z = \frac{I_x + I_y}{2}\)
Correct Answer: \(I_z = I_x + I_y\)
Explanation: According to the perpendicular axis theorem, if a lamina lies in the XY-plane, then the moment of inertia about the perpendicular axis \(Z\) is the sum of the moments of inertia about the X and Y axes in the plane.
413. In the theorem \(I_z = I_x + I_y\), what does \(I_z\) represent?
ⓐ. Moment of inertia about the x-axis
ⓑ. Moment of inertia about the y-axis
ⓒ. Moment of inertia about the axis perpendicular to the plane of the lamina
ⓓ. Moment of inertia about the diagonal axis of the lamina
Correct Answer: Moment of inertia about the axis perpendicular to the plane of the lamina
Explanation: \(I_z\) is the moment of inertia of the lamina about the perpendicular axis, which is normal to the plane in which the body lies.
414. Which of the following steps is important in the proof of the perpendicular axis theorem?
ⓐ. Expanding the equation \((x+d)^2\)
ⓑ. Resolving the distance of particles into x and y coordinates
ⓒ. Using Hooke’s law
ⓓ. Assuming torque is zero
Correct Answer: Resolving the distance of particles into x and y coordinates
Explanation: In the proof, the perpendicular distance from the Z-axis is expressed as \(r^2 = x^2 + y^2\). This leads to \(I_z = \sum m_i(x_i^2 + y_i^2) = I_x + I_y\).
415. The perpendicular axis theorem cannot be applied to:
ⓐ. A uniform disc
ⓑ. A thin rectangular plate
ⓒ. A solid sphere
ⓓ. A circular lamina
Correct Answer: A solid sphere
Explanation: The theorem applies only to planar laminae. Since a sphere is three-dimensional, the perpendicular axis theorem cannot be applied.
416. The perpendicular axis theorem is derived from:
ⓐ. The definition of torque
ⓑ. The definition of angular velocity
ⓒ. The definition of moment of inertia for point masses
ⓓ. Newton’s second law of motion
Correct Answer: The definition of moment of inertia for point masses
Explanation: The proof uses \(I = \sum m_i r_i^2\), where \(r^2 = x^2 + y^2\). Substituting gives the relation \(I_z = I_x + I_y\).
417. For a uniform circular lamina of radius \(R\), which statement is correct using the perpendicular axis theorem?
ⓐ. \(I_z = 2I_x\)
ⓑ. \(I_z = \frac{1}{2}I_x\)
ⓒ. \(I_z = I_x – I_y\)
ⓓ. \(I_z = 0\)
Correct Answer: \(I_z = 2I_x\)
Explanation: For a circular lamina, by symmetry \(I_x = I_y\). Hence, \(I_z = I_x + I_y = 2I_x\).
418. The perpendicular axis theorem is particularly useful in finding the moment of inertia of:
ⓐ. Rods about their ends
ⓑ. Discs about diameters and perpendicular axes
ⓒ. Cylinders about their central axis
ⓓ. Spheres about diameters
Correct Answer: Discs about diameters and perpendicular axes
Explanation: For discs or laminae, once the MOI about a diameter is known, the perpendicular axis theorem helps quickly calculate the MOI about the perpendicular axis.
419. What is the moment of inertia of a uniform circular disc of mass \(M\) and radius \(R\) about an axis perpendicular to the plane and passing through its center?
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(\frac{1}{4}MR^2\)
ⓒ. \(MR^2\)
ⓓ. \(\frac{3}{4}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: Using the perpendicular axis theorem, \(I_z = I_x + I_y\). For a disc, \(I_x = I_y = \frac{1}{4}MR^2\). Therefore, \(I_z = \frac{1}{2}MR^2\).
420. Why is the perpendicular axis theorem restricted to planar laminae only?
ⓐ. Because planar laminae have uniform density
ⓑ. Because in planar laminae, all mass elements lie in a single plane
ⓒ. Because 3D bodies have zero moment of inertia
ⓓ. Because torque is absent in laminae
Correct Answer: Because in planar laminae, all mass elements lie in a single plane
Explanation: The theorem requires that all particles lie in the same plane, so the perpendicular distance from the Z-axis can be resolved as \(r^2 = x^2 + y^2\). This condition is not satisfied in 3D bodies.
421. Using the parallel axis theorem, what is the moment of inertia of a thin uniform rod of mass \(M\) and length \(L\) about an axis perpendicular to the rod and passing through one end?
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{2}ML^2\)
ⓓ. \(ML^2\)
Correct Answer: \(\frac{1}{3}ML^2\)
Explanation: The moment of inertia about the center is \(I_{\text{cm}} = \frac{1}{12}ML^2\). By parallel axis theorem, shifting the axis by \(d = \frac{L}{2}\): \(I = I_{\text{cm}} + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2\).
422. A uniform disc of mass \(M\) and radius \(R\) has its moment of inertia about a diameter equal to:
ⓐ. \(\frac{1}{4}MR^2\)
ⓑ. \(\frac{1}{2}MR^2\)
ⓒ. \(\frac{3}{4}MR^2\)
ⓓ. \(MR^2\)
Correct Answer: \(\frac{1}{4}MR^2\)
Explanation: Using the perpendicular axis theorem: \(I_z = I_x + I_y = \frac{1}{2}MR^2\). By symmetry, \(I_x = I_y\). Thus, \(I_x = I_y = \frac{1}{4}MR^2\).
423. What is the moment of inertia of a circular ring of mass \(M\) and radius \(R\) about a tangent in the plane of the ring?
ⓐ. \(MR^2\)
ⓑ. \(2MR^2\)
ⓒ. \(\frac{3}{2}MR^2\)
ⓓ. \(\frac{1}{2}MR^2\)
Correct Answer: \(2MR^2\)
Explanation: About its center and perpendicular axis, \(I_z = MR^2\). About a tangent in the plane, use the parallel axis theorem: \(I = I_z + MR^2 = 2MR^2\).
424. The moment of inertia of a uniform solid sphere about any diameter is:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(\frac{2}{5}MR^2\)
ⓒ. \(\frac{3}{5}MR^2\)
ⓓ. \(\frac{7}{5}MR^2\)
Correct Answer: \(\frac{2}{5}MR^2\)
Explanation: For a uniform solid sphere, the standard result derived by integration is \(I = \frac{2}{5}MR^2\) about any diameter.
425. For a thin rectangular lamina of mass \(M\), length \(l\), and breadth \(b\), the moment of inertia about an axis perpendicular to the plane and passing through the center is:
ⓐ. \(\frac{1}{12}M(l^2 + b^2)\)
ⓑ. \(\frac{1}{6}M(l^2 + b^2)\)
ⓒ. \(\frac{1}{4}M(l^2 + b^2)\)
ⓓ. \(\frac{1}{2}M(l^2 + b^2)\)
Correct Answer: \(\frac{1}{12}M(l^2 + b^2)\)
Explanation: Using perpendicular axis theorem, \(I_z = I_x + I_y\), with \(I_x = \frac{1}{12}Ml^2\) and \(I_y = \frac{1}{12}Mb^2\). Thus, \(I_z = \frac{1}{12}M(l^2+b^2)\).
426. Using parallel axis theorem, what is the moment of inertia of a disc of mass \(M\) and radius \(R\) about an axis tangent to the disc and lying in its plane?
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{3}{2}MR^2\)
ⓓ. \(2MR^2\)
Correct Answer: \(\frac{3}{2}MR^2\)
Explanation: About its central axis in-plane (diameter), \(I = \frac{1}{4}MR^2\). Using parallel axis theorem with \(d = R\): \(I = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2\). Correction: Actually, for tangent in plane, correct calculation is \(I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2\).
427. The moment of inertia of a thin rod of length \(L\) and mass \(M\) about an axis perpendicular to its length and passing through its midpoint is:
ⓐ. \(\frac{1}{12}ML^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{2}ML^2\)
ⓓ. \(ML^2\)
Correct Answer: \(\frac{1}{12}ML^2\)
Explanation: Standard formula gives \(I = \frac{1}{12}ML^2\) for a thin rod about its midpoint and perpendicular axis.
428. The perpendicular axis theorem can be used to calculate the moment of inertia of a thin circular disc about its center and perpendicular axis because:
ⓐ. The disc is a 3D body
ⓑ. The disc is a 2D lamina
ⓒ. The disc has rotational symmetry
ⓓ. The disc has zero torque
Correct Answer: The disc is a 2D lamina
Explanation: The theorem applies only to laminae. Since the disc is a flat 2D object, the theorem holds and gives \(I_z = I_x + I_y\).
429. A solid cylinder of mass \(M\) and radius \(R\) has a moment of inertia about its axis equal to:
ⓐ. \(\frac{1}{2}MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{3}{2}MR^2\)
ⓓ. \(\frac{2}{5}MR^2\)
Correct Answer: \(\frac{1}{2}MR^2\)
Explanation: For a solid cylinder rotating about its own axis, the moment of inertia is \(\frac{1}{2}MR^2\).
430. If a body is composed of multiple simple shapes, the total moment of inertia is calculated by:
ⓐ. Subtracting the moments of inertia of each shape
ⓑ. Adding the moments of inertia of each shape
ⓒ. Multiplying the moments of inertia of each shape
ⓓ. Dividing the largest by the smallest
Correct Answer: Adding the moments of inertia of each shape
Explanation: For composite bodies, the total moment of inertia is the sum of the individual moments of inertia of each part about the same axis, considering parallel and perpendicular axis theorems if needed.
431. What is the angular form of Newton’s second law for rotational motion about a fixed axis?
ⓐ. \(\tau = I \alpha\)
ⓑ. \(F = ma\)
ⓒ. \(p = mv\)
ⓓ. \(L = I \omega\)
Correct Answer: \(\tau = I \alpha\)
Explanation: The rotational analogue of Newton’s second law states that the net torque \(\tau\) acting on a rigid body about a fixed axis is equal to the product of the moment of inertia \(I\) and angular acceleration \(\alpha\).
432. In the equation \(\tau = I \alpha\), what does \(\alpha\) represent?
ⓐ. Linear acceleration of the body
ⓑ. Angular acceleration of the body
ⓒ. Linear velocity of the body
ⓓ. Angular momentum of the body
Correct Answer: Angular acceleration of the body
Explanation: \(\alpha\) is the rate of change of angular velocity with respect to time, and it represents how quickly the rotational speed of the body changes.
433. Which quantity in rotational motion is analogous to force in linear motion?
ⓐ. Angular velocity
ⓑ. Torque
ⓒ. Angular momentum
ⓓ. Moment of inertia
Correct Answer: Torque
Explanation: Just as force changes linear motion, torque is the rotational equivalent that changes angular motion. It causes angular acceleration when applied to a rigid body.
434. Which quantity in rotational motion is analogous to mass in linear motion?
ⓐ. Torque
ⓑ. Angular velocity
ⓒ. Moment of inertia
ⓓ. Angular momentum
Correct Answer: Moment of inertia
Explanation: Moment of inertia is the rotational analogue of mass. It represents the resistance of a body to changes in its angular motion.
435. If the torque on a rotating body is doubled, what happens to the angular acceleration (assuming moment of inertia is constant)?
ⓐ. It becomes half
ⓑ. It doubles
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It doubles
Explanation: From \(\tau = I \alpha\), angular acceleration is directly proportional to torque. So, doubling torque doubles angular acceleration if \(I\) remains the same.
436. If a flywheel of moment of inertia \(10 \, \text{kg m}^2\) is subjected to a torque of \(50 \, \text{Nm}\), what is its angular acceleration?
ⓐ. \(0.5 \, \text{rad/s}^2\)
ⓑ. \(2 \, \text{rad/s}^2\)
ⓒ. \(5 \, \text{rad/s}^2\)
ⓓ. \(10 \, \text{rad/s}^2\)
Correct Answer: \(5 \, \text{rad/s}^2\)
Explanation: Using \(\alpha = \tau/I\), we get \(\alpha = 50 / 10 = 5 \, \text{rad/s}^2\).
437. What condition must be true for the equation \(\tau = I \alpha\) to hold?
ⓐ. Axis of rotation must pass through center of mass
ⓑ. Axis of rotation must be fixed
ⓒ. Torque must always be zero
ⓓ. Angular velocity must be constant
Correct Answer: Axis of rotation must be fixed
Explanation: The angular form of Newton’s law is derived for rigid bodies rotating about a fixed axis, where the distribution of mass relative to the axis does not change.
438. What is the work-energy relation in rotational motion analogous to \(W = F \cdot d\) in linear motion?
ⓐ. \(W = \tau \cdot \theta\)
ⓑ. \(W = I \cdot \alpha\)
ⓒ. \(W = \omega \cdot t\)
ⓓ. \(W = L \cdot \omega\)
Correct Answer: \(W = \tau \cdot \theta\)
Explanation: In rotational dynamics, work done by torque is equal to torque multiplied by the angular displacement of the body.
439. In rotational motion, angular momentum \(L\) is related to torque \(\tau\) by which expression?
ⓐ. \(\tau = \frac{dp}{dt}\)
ⓑ. \(\tau = \frac{dL}{dt}\)
ⓒ. \(\tau = L \cdot \omega\)
ⓓ. \(\tau = I \cdot \omega\)
Correct Answer: \(\tau = \frac{dL}{dt}\)
Explanation: Just like force is the rate of change of linear momentum, torque is the rate of change of angular momentum.
440. Which physical principle connects linear motion with rotational motion in Newton’s second law?
ⓐ. Equivalence principle
ⓑ. Work-energy theorem
ⓒ. Analogy between force and torque, mass and moment of inertia
ⓓ. Law of gravitation
Correct Answer: Analogy between force and torque, mass and moment of inertia
Explanation: Newton’s second law has a direct analogy in rotational motion where force ↔ torque, mass ↔ moment of inertia, and linear acceleration ↔ angular acceleration.
441. What is the mathematical relation between torque and angular acceleration for a rigid body?
ⓐ. \(\tau = I \omega\)
ⓑ. \(\tau = I \alpha\)
ⓒ. \(\tau = m a\)
ⓓ. \(\tau = \frac{dL}{dt}\)
Correct Answer: \(\tau = I \alpha\)
Explanation: Torque is directly proportional to angular acceleration, with the moment of inertia \(I\) as the proportionality constant. This is the angular form of Newton’s second law for rotation.
442. If the torque acting on a wheel is zero, what will be the angular acceleration of the wheel?
ⓐ. Infinite
ⓑ. Constant but non-zero
ⓒ. Zero
ⓓ. Cannot be determined
Correct Answer: Zero
Explanation: Since \(\tau = I \alpha\), if torque is zero, then angular acceleration must also be zero. The wheel will either remain at rest or continue to rotate with constant angular velocity.
443. A disc of moment of inertia \(2 \, \text{kg·m}^2\) experiences a torque of \(10 \, \text{N·m}\). What is its angular acceleration?
ⓐ. \(2 \, \text{rad/s}^2\)
ⓑ. \(5 \, \text{rad/s}^2\)
ⓒ. \(10 \, \text{rad/s}^2\)
ⓓ. \(20 \, \text{rad/s}^2\)
Correct Answer: \(5 \, \text{rad/s}^2\)
Explanation: Using \(\alpha = \tau / I\), we get \(\alpha = 10 / 2 = 5 \, \text{rad/s}^2\).
444. If the torque on a rotating rigid body is doubled, how does the angular acceleration change (keeping moment of inertia constant)?
ⓐ. It becomes half
ⓑ. It doubles
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It doubles
Explanation: Angular acceleration is directly proportional to torque when moment of inertia is constant. Thus, doubling torque doubles angular acceleration.
445. Why does a heavier flywheel have a smaller angular acceleration for the same applied torque compared to a lighter one?
ⓐ. Because torque is inversely proportional to angular momentum
ⓑ. Because it has a larger moment of inertia
ⓒ. Because torque decreases with mass
ⓓ. Because angular velocity remains constant
Correct Answer: Because it has a larger moment of inertia
Explanation: Angular acceleration is \(\alpha = \tau / I\). A heavier flywheel has larger \(I\), which reduces angular acceleration for the same torque.
446. A torque of \(15 \, \text{N·m}\) produces an angular acceleration of \(3 \, \text{rad/s}^2\) in a wheel. What is the wheel’s moment of inertia?
ⓐ. \(3 \, \text{kg·m}^2\)
ⓑ. \(4 \, \text{kg·m}^2\)
ⓒ. \(5 \, \text{kg·m}^2\)
ⓓ. \(6 \, \text{kg·m}^2\)
Correct Answer: \(5 \, \text{kg·m}^2\)
Explanation: From \(I = \tau / \alpha\), \(I = 15 / 3 = 5 \, \text{kg·m}^2\).
447. Which of the following best explains why torque is necessary to change rotational motion?
ⓐ. Torque increases mass of the object
ⓑ. Torque changes linear momentum directly
ⓒ. Torque produces angular acceleration proportional to moment of inertia
ⓓ. Torque decreases angular displacement
Correct Answer: Torque produces angular acceleration proportional to moment of inertia
Explanation: Just as force changes linear velocity, torque changes angular velocity by producing angular acceleration in proportion to the body’s resistance, i.e., moment of inertia.
448. If a wheel has a moment of inertia of \(8 \, \text{kg·m}^2\) and is subjected to a torque of \(16 \, \text{N·m}\), its angular acceleration is:
ⓐ. \(1 \, \text{rad/s}^2\)
ⓑ. \(2 \, \text{rad/s}^2\)
ⓒ. \(4 \, \text{rad/s}^2\)
ⓓ. \(8 \, \text{rad/s}^2\)
Correct Answer: \(2 \, \text{rad/s}^2\)
Explanation: Angular acceleration is \(\alpha = \tau/I = 16/8 = 2 \, \text{rad/s}^2\).
449. When does the relation \(\tau = I \alpha\) not hold true?
ⓐ. When torque is not constant
ⓑ. When the axis of rotation is not fixed
ⓒ. When the body is rigid
ⓓ. When angular acceleration is zero
Correct Answer: When the axis of rotation is not fixed
Explanation: The relation applies only when the axis of rotation is fixed. If the axis shifts or moves, additional terms must be considered for motion of the center of mass.
450. Why is torque considered the rotational analogue of force?
ⓐ. Both depend on linear displacement
ⓑ. Both depend on moment of inertia
ⓒ. Both cause acceleration (linear or angular) in proportion to resistance
ⓓ. Both are measured in joules
Correct Answer: Both cause acceleration (linear or angular) in proportion to resistance
Explanation: Force changes linear motion by producing linear acceleration proportional to mass, while torque changes rotational motion by producing angular acceleration proportional to moment of inertia.
451. What is the expression for rotational kinetic energy of a rigid body rotating with angular velocity \(\omega\)?
ⓐ. \(KE = \frac{1}{2} I \omega^2\)
ⓑ. \(KE = \frac{1}{2} mv^2\)
ⓒ. \(KE = I \alpha^2\)
ⓓ. \(KE = \tau \cdot \theta\)
Correct Answer: \(KE = \frac{1}{2} I \omega^2\)
Explanation: The rotational kinetic energy of a rigid body is analogous to translational kinetic energy. Here, \(I\) plays the role of mass and \(\omega\) the role of velocity.
452. The work-energy theorem in rotational motion can be expressed as:
ⓐ. Work done = Change in potential energy
ⓑ. Work done = Change in angular momentum
ⓒ. Work done by torque = Change in rotational kinetic energy
ⓓ. Work done = Torque × Angular velocity
Correct Answer: Work done by torque = Change in rotational kinetic energy
Explanation: The theorem states that the work done by external torques on a rotating body equals the change in its rotational kinetic energy.
453. If a wheel of moment of inertia \(I = 4 \, \text{kg·m}^2\) rotates with angular velocity \(10 \, \text{rad/s}\), what is its rotational kinetic energy?
ⓐ. \(100 \, \text{J}\)
ⓑ. \(200 \, \text{J}\)
ⓒ. \(250 \, \text{J}\)
ⓓ. \(400 \, \text{J}\)
Correct Answer: \(200 \, \text{J}\)
Explanation: Using \(KE = \tfrac{1}{2} I \omega^2 = \tfrac{1}{2} \times 4 \times 100 = 200 \, \text{J}\).
454. Which quantity in rotational motion is analogous to displacement in linear motion when calculating work?
ⓐ. Angular velocity
ⓑ. Angular momentum
ⓒ. Angular displacement
ⓓ. Moment of inertia
Correct Answer: Angular displacement
Explanation: In rotational motion, work done is given by \(W = \tau \cdot \theta\), where \(\theta\) (angular displacement) is analogous to linear displacement in translation.
455. If the torque acting on a body is constant, the work done by the torque after an angular displacement \(\theta\) is:
ⓐ. \(W = \frac{1}{2} I \omega^2\)
ⓑ. \(W = \tau \theta\)
ⓒ. \(W = I \alpha \theta\)
ⓓ. \(W = L \omega\)
Correct Answer: \(W = \tau \theta\)
Explanation: For constant torque, work done equals torque multiplied by angular displacement, analogous to \(W = F \cdot d\) in linear motion.
456. A disc of moment of inertia \(2 \, \text{kg·m}^2\) starts from rest and is subjected to a constant torque of \(4 \, \text{N·m}\). What is its rotational kinetic energy after rotating through \(2 \, \text{rad}\)?
ⓐ. \(4 \, \text{J}\)
ⓑ. \(6 \, \text{J}\)
ⓒ. \(8 \, \text{J}\)
ⓓ. \(10 \, \text{J}\)
Correct Answer: \(8 \, \text{J}\)
Explanation: Work done by torque \(W = \tau \cdot \theta = 4 \times 2 = 8 \, \text{J}\). This equals the rotational kinetic energy gained by the disc.
457. Which of the following statements is true about rotational kinetic energy?
ⓐ. It depends only on torque
ⓑ. It depends only on angular velocity
ⓒ. It depends on both moment of inertia and angular velocity
ⓓ. It is independent of axis of rotation
Correct Answer: It depends on both moment of inertia and angular velocity
Explanation: \(KE = \tfrac{1}{2} I \omega^2\), showing dependence on both inertia and angular speed.
458. If angular velocity of a rotating object is doubled, its rotational kinetic energy becomes:
ⓐ. Half
ⓑ. Double
ⓒ. Four times
ⓓ. Eight times
Correct Answer: Four times
Explanation: Since \(KE \propto \omega^2\), doubling angular velocity increases kinetic energy by a factor of four.
459. The work-energy theorem in rotational motion is a direct consequence of:
ⓐ. Newton’s third law
ⓑ. Torque-angular acceleration relationship
ⓒ. Conservation of momentum
ⓓ. Law of gravitation
Correct Answer: Torque-angular acceleration relationship
Explanation: The theorem follows from integrating the relation \(\tau = I\alpha\) over angular displacement, linking torque, work, and change in kinetic energy.
460. Which of the following is an example where rotational kinetic energy plays a crucial role?
ⓐ. A car moving on a straight road
ⓑ. A pendulum oscillating
ⓒ. A rolling wheel of a bicycle
ⓓ. A stretched spring
Correct Answer: A rolling wheel of a bicycle
Explanation: A rolling wheel possesses both translational and rotational kinetic energy. The rotational part is given by \(\tfrac{1}{2} I \omega^2\), which contributes significantly to the total energy of motion.
461. What is the definition of angular momentum for a particle rotating about a fixed axis?
ⓐ. \(L = m v r\)
ⓑ. \(L = I \alpha\)
ⓒ. \(L = I \omega\)
ⓓ. \(L = \tau \theta\)
Correct Answer: \(L = I \omega\)
Explanation: For rotation about a fixed axis, angular momentum is given by \(L = I \omega\), where \(I\) is the moment of inertia about the axis and \(\omega\) is the angular velocity.
462. Which physical quantity in linear motion is analogous to angular momentum in rotational motion?
ⓐ. Kinetic energy
ⓑ. Linear momentum
ⓒ. Torque
ⓓ. Displacement
Correct Answer: Linear momentum
Explanation: Angular momentum is the rotational analogue of linear momentum. Just as \(p = mv\), angular momentum is \(L = I\omega\).
463. What are the SI units of angular momentum?
ⓐ. \(\text{kg·m}^2/\text{s}^2\)
ⓑ. \(\text{kg·m}/\text{s}\)
ⓒ. \(\text{kg·m}^2/\text{s}\)
ⓓ. \(\text{N·m}\)
Correct Answer: \(\text{kg·m}^2/\text{s}\)
Explanation: Angular momentum has dimensions of moment of inertia (\(\text{kg·m}^2\)) times angular velocity (\(1/\text{s}\)), giving \(\text{kg·m}^2/\text{s}\).
464. If a disc has a moment of inertia of \(4 \, \text{kg·m}^2\) and is rotating with angular velocity \(3 \, \text{rad/s}\), what is its angular momentum?
ⓐ. \(6 \, \text{kg·m}^2/\text{s}\)
ⓑ. \(8 \, \text{kg·m}^2/\text{s}\)
ⓒ. \(10 \, \text{kg·m}^2/\text{s}\)
ⓓ. \(12 \, \text{kg·m}^2/\text{s}\)
Correct Answer: \(12 \, \text{kg·m}^2/\text{s}\)
Explanation: Using \(L = I \omega\), we get \(L = 4 \times 3 = 12 \, \text{kg·m}^2/\text{s}\).
465. Why is angular momentum considered a vector quantity?
ⓐ. Because it depends only on mass
ⓑ. Because it has both magnitude and a direction given by the axis of rotation
ⓒ. Because it depends only on angular velocity
ⓓ. Because it is always conserved
Correct Answer: Because it has both magnitude and a direction given by the axis of rotation
Explanation: Angular momentum has magnitude \(L = I \omega\) and direction along the axis of rotation, determined by the right-hand rule, making it a vector.
466. In the case of a rigid body rotating about a fixed axis, angular momentum is always directed:
ⓐ. Along the axis of rotation
ⓑ. Perpendicular to the axis of rotation
ⓒ. Tangential to the path of particles
ⓓ. In the direction of angular displacement
Correct Answer: Along the axis of rotation
Explanation: For a fixed axis, the angular momentum vector is aligned with the axis of rotation, following the right-hand rule.
467. If angular velocity is doubled while keeping moment of inertia constant, how does angular momentum change?
ⓐ. It remains the same
ⓑ. It doubles
ⓒ. It becomes four times
ⓓ. It becomes half
Correct Answer: It doubles
Explanation: Angular momentum is \(L = I \omega\). With constant \(I\), doubling \(\omega\) doubles \(L\).
468. For a body of moment of inertia \(I\), what happens to angular momentum if the axis of rotation changes?
ⓐ. It always remains the same
ⓑ. It always doubles
ⓒ. It changes because moment of inertia depends on axis
ⓓ. It becomes zero
Correct Answer: It changes because moment of inertia depends on axis
Explanation: Angular momentum depends on both \(I\) and \(\omega\). Changing the axis changes \(I\), hence altering angular momentum.
469. Which law connects torque and angular momentum?
ⓐ. \(\tau = I \alpha\)
ⓑ. \(\tau = \frac{dL}{dt}\)
ⓒ. \(\tau = L \cdot \omega\)
ⓓ. \(\tau = F \cdot r\)
Correct Answer: \(\tau = \frac{dL}{dt}\)
Explanation: The time rate of change of angular momentum equals the torque acting on the body, analogous to \(F = dp/dt\) in linear motion.
470. Why is angular momentum important in analyzing rotational systems?
ⓐ. Because it replaces linear velocity
ⓑ. Because it provides conservation laws useful in solving problems
ⓒ. Because it is easier to calculate than torque
ⓓ. Because it does not depend on moment of inertia
Correct Answer: Because it provides conservation laws useful in solving problems
Explanation: Angular momentum is conserved in the absence of external torque, making it fundamental in analyzing rotational systems like planetary motion, gyroscopes, and collisions.
471. Which principle explains why a rotating neutron star (pulsar) spins faster as it collapses?
ⓐ. Conservation of energy
ⓑ. Conservation of torque
ⓒ. Conservation of angular momentum
ⓓ. Conservation of mass
Correct Answer: Conservation of angular momentum
Explanation: As the star collapses, its radius decreases, reducing its moment of inertia. Since angular momentum is conserved, angular velocity increases, making the star spin faster.
472. Why do engineers design flywheels with large moments of inertia?
ⓐ. To reduce weight
ⓑ. To store large amounts of rotational kinetic energy
ⓒ. To increase torque directly
ⓓ. To decrease angular velocity
Correct Answer: To store large amounts of rotational kinetic energy
Explanation: Flywheels with large moment of inertia can store energy efficiently as \(KE = \tfrac{1}{2} I \omega^2\). This stored energy can be used to smooth power output in engines or provide backup power.
473. The stability of artificial satellites in orbit is maintained using which principle?
ⓐ. Conservation of energy
ⓑ. Newton’s first law
ⓒ. Conservation of angular momentum
ⓓ. Universal law of gravitation
Correct Answer: Conservation of angular momentum
Explanation: Satellites use conservation of angular momentum for orientation and stability. Gyroscopes and reaction wheels apply this principle to control attitude without external torque.
474. Why do galaxies rotate with a nearly constant angular momentum distribution?
ⓐ. Due to conservation of mass
ⓑ. Due to gravitational attraction only
ⓒ. Due to conservation of angular momentum of stars and gas clouds
ⓓ. Due to torque applied by dark matter
Correct Answer: Due to conservation of angular momentum of stars and gas clouds
Explanation: As galaxies form and evolve, stars and gas clouds redistribute mass but conserve angular momentum, leading to stable rotational patterns observed in galaxies.
475. Which engineering device uses conservation of angular momentum for navigation without GPS?
ⓐ. Accelerometer
ⓑ. Gyroscope
ⓒ. Barometer
ⓓ. Dynamometer
Correct Answer: Gyroscope
Explanation: Gyroscopes use conservation of angular momentum to maintain orientation. They are essential in aircraft, ships, spacecraft, and submarines for navigation without relying on external signals.
476. Why do astronauts use rotating chairs or platforms during training?
ⓐ. To practice linear acceleration
ⓑ. To simulate conservation of angular momentum
ⓒ. To reduce gravitational effects
ⓓ. To measure torque directly
Correct Answer: To simulate conservation of angular momentum
Explanation: When astronauts pull in their arms or legs, they spin faster due to reduced moment of inertia while angular momentum is conserved. This helps them practice body control in space.
477. The phenomenon of accretion disks around black holes is strongly influenced by:
ⓐ. Linear momentum conservation
ⓑ. Torque due to magnetic fields only
ⓒ. Conservation of angular momentum of matter spiraling inward
ⓓ. Conservation of energy alone
Correct Answer: Conservation of angular momentum of matter spiraling inward
Explanation: Gas and dust falling into black holes form accretion disks because they conserve angular momentum, causing the matter to spiral around instead of falling directly inward.
478. Why are reaction wheels used in spacecraft?
ⓐ. To reduce gravitational pull of Earth
ⓑ. To control rotation using conservation of angular momentum
ⓒ. To increase orbital velocity directly
ⓓ. To measure energy consumption
Correct Answer: To control rotation using conservation of angular momentum
Explanation: Reaction wheels spin inside spacecraft, and by conservation of angular momentum, they allow precise control of spacecraft orientation without fuel consumption.
479. Why do engineers consider angular momentum in the design of turbines and rotors?
ⓐ. To reduce torque losses
ⓑ. To improve linear displacement
ⓒ. To ensure stability and efficient energy transfer
ⓓ. To reduce the number of blades required
Correct Answer: To ensure stability and efficient energy transfer
Explanation: Turbines and rotors involve high rotational speeds. Conservation of angular momentum ensures smooth energy transfer and stability in power plants and aircraft engines.
480. In astrophysics, why do planets in our solar system continue to orbit the Sun without slowing down significantly?
ⓐ. Because torque applied by the Sun keeps them rotating
ⓑ. Because their angular momentum is conserved in the absence of external torque
ⓒ. Because their mass keeps increasing
ⓓ. Because gravitational force increases with time
Correct Answer: Because their angular momentum is conserved in the absence of external torque
Explanation: In the near absence of external torque, planetary systems conserve angular momentum, allowing planets to maintain nearly constant orbital motion around the Sun for billions of years.
481. What is the primary condition for rolling motion without slipping?
ⓐ. \(v = r \alpha\)
ⓑ. \(v = \omega r\)
ⓒ. \(a = r \omega\)
ⓓ. \(\tau = I \omega\)
Correct Answer: \(v = \omega r\)
Explanation: For pure rolling motion without slipping, the velocity of the center of mass \(v\) must equal the product of angular velocity \(\omega\) and radius \(r\). This ensures no relative motion at the point of contact.
482. Which force ensures rolling motion without slipping on a horizontal surface?
ⓐ. Normal reaction
ⓑ. Gravitational force
ⓒ. Frictional force
ⓓ. Centripetal force
Correct Answer: Frictional force
Explanation: Friction prevents slipping and provides the necessary torque for rolling. Without sufficient friction, the body would slide instead of rolling.
483. For a rolling sphere, the point of contact with the ground has what velocity relative to the ground?
ⓐ. \(v\)
ⓑ. \(\omega r\)
ⓒ. Zero
ⓓ. \(2v\)
Correct Answer: Zero
Explanation: In rolling without slipping, the bottom point is instantaneously at rest relative to the ground, making its velocity zero.
484. A solid sphere of radius \(r\) is rolling without slipping with angular velocity \(\omega\). The velocity of its center of mass is:
ⓐ. \(\frac{\omega}{r}\)
ⓑ. \(\omega r\)
ⓒ. \(\frac{r}{\omega}\)
ⓓ. \(\omega^2 r\)
Correct Answer: \(\omega r\)
Explanation: The velocity of the center of mass in pure rolling is \(v_{CM} = \omega r\).
485. Which of the following must hold true for rolling without slipping?
ⓐ. Kinetic energy is only rotational
ⓑ. Linear velocity of CM equals tangential velocity at the rim
ⓒ. Angular velocity is zero
ⓓ. Acceleration of CM equals torque
Correct Answer: Linear velocity of CM equals tangential velocity at the rim
Explanation: For rolling without slipping, the tangential velocity at the edge equals the linear velocity of the center of mass, ensuring no relative slipping.
486. When a cylinder rolls without slipping, what is the acceleration condition?
ⓐ. \(a_{CM} = \omega r\)
ⓑ. \(a_{CM} = \alpha r\)
ⓒ. \(a_{CM} = \frac{\tau}{r}\)
ⓓ. \(a_{CM} = \frac{I}{mr}\)
Correct Answer: \(a_{CM} = \alpha r\)
Explanation: For rolling without slipping, the linear acceleration of the center of mass equals the angular acceleration times the radius.
487. In rolling without slipping, the instantaneous axis of rotation of the rolling body is located at:
ⓐ. The center of mass
ⓑ. The axis perpendicular to motion
ⓒ. The point of contact with the surface
ⓓ. At infinity
Correct Answer: The point of contact with the surface
Explanation: In rolling without slipping, the point of contact is instantaneously at rest and acts as the instantaneous axis of rotation.
488. A wheel of radius 0.5 m rolls without slipping with angular speed \(2 \, \text{rad/s}\). What is the speed of its center of mass?
ⓐ. \(0.5 \, \text{m/s}\)
ⓑ. \(1 \, \text{m/s}\)
ⓒ. \(2 \, \text{m/s}\)
ⓓ. \(4 \, \text{m/s}\)
Correct Answer: \(1 \, \text{m/s}\)
Explanation: \(v = \omega r = 2 \times 0.5 = 1 \, \text{m/s}\).
489. Which type of friction is responsible for maintaining rolling without slipping?
ⓐ. Kinetic friction
ⓑ. Static friction
ⓒ. Rolling friction
ⓓ. Viscous friction
Correct Answer: Static friction
Explanation: Rolling without slipping requires static friction, which ensures no relative motion at the point of contact between the surface and wheel.
490. If a rolling body begins to slip, which of the following relationships no longer holds?
ⓐ. \(v = \omega r\)
ⓑ. \(a = \alpha r\)
ⓒ. Point of contact is instantaneously at rest
ⓓ. All of the above
Correct Answer: All of the above
Explanation: When slipping occurs, the pure rolling conditions (\(v = \omega r\), \(a = \alpha r\), and rest at point of contact) break down, leading to sliding motion.
491. Which condition correctly defines the kinematics of rolling motion without slipping?
ⓐ. \(v_{CM} = \alpha r\)
ⓑ. \(v_{CM} = \omega r\)
ⓒ. \(v_{CM} = \tau r\)
ⓓ. \(v_{CM} = I \omega\)
Correct Answer: \(v_{CM} = \omega r\)
Explanation: For rolling without slipping, the velocity of the center of mass equals the product of angular velocity and radius, ensuring no relative slipping at the contact point.
492. In rolling motion, the acceleration of the center of mass is related to angular acceleration by:
ⓐ. \(a_{CM} = \tau / r\)
ⓑ. \(a_{CM} = \alpha r\)
ⓒ. \(a_{CM} = \omega / r\)
ⓓ. \(a_{CM} = I \alpha\)
Correct Answer: \(a_{CM} = \alpha r\)
Explanation: In pure rolling, linear acceleration of the center of mass is proportional to angular acceleration with the factor of radius \(r\).
493. For a wheel rolling without slipping, what is the velocity of the topmost point relative to the ground?
ⓐ. Zero
ⓑ. \(v_{CM}\)
ⓒ. \(2v_{CM}\)
ⓓ. \(\omega r/2\)
Correct Answer: \(2v_{CM}\)
Explanation: The topmost point has velocity equal to \(v_{CM} + \omega r = 2v_{CM}\), since both translational and rotational velocities add in the same direction.
494. For a rolling cylinder, what is the acceleration of the point of contact with the ground relative to the ground?
ⓐ. Zero
ⓑ. \(a_{CM}\)
ⓒ. \(2a_{CM}\)
ⓓ. \(\alpha r\)
Correct Answer: Zero
Explanation: The point of contact is instantaneously at rest in pure rolling, so its acceleration relative to the ground is zero.
495. Which of the following best describes the motion of a point on the rim of a rolling wheel?
ⓐ. Straight line motion
ⓑ. Pure rotation only
ⓒ. Translation only
ⓓ. Cycloidal path
Correct Answer: Cycloidal path
Explanation: A point on the rim of a rolling wheel traces a cycloid, resulting from the combination of rotational and translational motion.
496. If the linear velocity of the center of mass is 3 m/s and the wheel radius is 0.5 m, what is the angular velocity for rolling without slipping?
ⓐ. 3 rad/s
ⓑ. 6 rad/s
ⓒ. 1.5 rad/s
ⓓ. 0.5 rad/s
Correct Answer: 6 rad/s
Explanation: Using \(v_{CM} = \omega r\), we get \(\omega = v_{CM}/r = 3/0.5 = 6 \, \text{rad/s}\).
497. What is the acceleration of the topmost point of a rolling wheel relative to the ground?
ⓐ. Zero
ⓑ. \(a_{CM}\)
ⓒ. \(2a_{CM}\)
ⓓ. \(\alpha r\)
Correct Answer: \(2a_{CM}\)
Explanation: The acceleration of the topmost point equals the sum of translational acceleration of the center of mass and tangential acceleration due to rotation, giving \(2a_{CM}\).
498. In rolling motion, the instantaneous axis of rotation lies at:
ⓐ. The center of the wheel
ⓑ. The topmost point of the wheel
ⓒ. The point of contact with the surface
ⓓ. At infinity
Correct Answer: The point of contact with the surface
Explanation: For rolling without slipping, the point of contact is instantaneously at rest and acts as the instantaneous axis of rotation.
499. For rolling without slipping, what is the velocity of the bottommost point of the wheel relative to the ground?
ⓐ. \(v_{CM}\)
ⓑ. Zero
ⓒ. \(2v_{CM}\)
ⓓ. \(\omega r\)
Correct Answer: Zero
Explanation: At the bottommost point, rotational velocity and translational velocity are equal and opposite, cancelling out to zero.
500. If a wheel is rolling with slipping, which of the following statements is true?
ⓐ. \(v_{CM} = \omega r\) holds true
ⓑ. The point of contact has zero velocity
ⓒ. The kinematic relationship \(v_{CM} = \omega r\) no longer holds
ⓓ. The wheel has no angular acceleration
Correct Answer: The kinematic relationship \(v_{CM} = \omega r\) no longer holds
Explanation: For rolling with slipping, the pure rolling condition breaks down, meaning \(v_{CM} \neq \omega r\), and the point of contact slides relative to the surface.
The chapter System of Particles and Rotational Motion is one of the most important chapters of Class 11 Physics (NCERT/CBSE syllabus), forming a foundation for higher-level mechanics. It is extremely relevant for board exams and also carries high weightage in competitive exams like JEE, NEET, and state-level entrance tests. This section focuses on questions related to rotational equilibrium, torque, angular momentum, and practical applications. Out of a total of 570 MCQs with answers, this page contains the fifth set of 100 MCQs with step-by-step explanations for better understanding.
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- 👉 This page contains: Fifth set of 100 solved MCQs.
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